Chapter 4.7, Problem 36E

To determine

To find: the minimum or maximum value of the function and describe domain and range of the function where the function is increasing or decreasing.

Answer to Problem 36E

Minimum value $\left(3,-147\right)$ , $D=\left\{x:x\in R\right\}$ and $R=\left\{y:y\in \left[-147,\infty \right)\right\}$ .

Explanation of Solution

Given:

  $f\left(x\right)=3\left(x-10\right)\left(x+4\right)$ .

Concept used:

The maximum value and minimum value of the function can be derived from slope of the function which is first derivative when it is equal zero it makes the extreme or critical value where slope is $0$ this point is either maxima or minima.

To check whether curve is Minima or maxima different again or take double differentiate if the value is greater than the point is minima and if the value come up to be less than $0$ then the point is maxima.

Mathematically:

  $\frac{d^{2}y}{d{x}^2}>0$ then there is a minima point exist in the graph.

  $\frac{d^{2}y}{d{x}^2}<0$ then the there is a maxima point exist in the graph.

Calculation:

The maximum value and minimum value of the function can be derived from slope of the function which is first derivative when it is equal zero it makes the extreme or critical value where slope is $0$ this point is either maxima or minima.

To check whether curve is minima or maxima different again or take double differentiate if the value is greater than the point is minima and if the value come up to be less than $0$ then the point is maxima.

Mathematically:

  $\frac{d^{2}y}{d{x}^2}>0$ then there is a minima point exist in the graph.

  $\frac{d^{2}y}{d{x}^2}<0$ then the there is a maxima point exist in the graph.

  $\begin{array}{l}f\left(x\right)=3\left(x-10\right)\left(x+4.\right)\\ f\left(x\right)=3\left(x^2+4x-10x-40\right).\\ f\left(x\right)=3x^2-18x-120.\end{array}$

First differentiate can be calculated as:

  $\begin{array}{l}{f}^{\prime }\left(x\right)=6x-18.\\ f^{″}\left(x\right)=6.\end{array}$

Here $f^{″}\left(x\right)=6>0$ .

The curve is open upward which is mean the point is minima and the minimum point is:

  $\begin{array}{l}{f}^{\prime }\left(x\right)=0.\\ 6x-18=0.\\ \Rightarrow x=3.\end{array}$

Therefore, $\left(3,-147\right)$ is the minimum value or point.

Domain of the function since it is independent the value can be whole real numbers.

Mathematically:

  $D=\left\{x:x\in R\right\}$ .

Range of the function can minimum value of square of any number is $0$ or square of number cannot be negative.

Range is taken from its minimum value to maximum value:

Mathematically:

  $R=\left\{y:y\in \left[-147,\infty \right)\right\}$ .

Hence, minimum value $\left(3,-147\right)$ , $D=\left\{x:x\in R\right\}$ and $R=\left\{y:y\in \left[-147,\infty \right)\right\}$ .